Last updated at March 11, 2017 by Teachoo

Transcript

Example 30 (Method 1) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane. The equation of a line passing through two points with position vectors 𝑎 & 𝑏 is 𝒓 = 𝒂 + 𝜆 ( 𝒃 − 𝒂) Given the line passes through the points ( 𝑏 − 𝑎) = (5 𝑖 + 1 𝑗 + 6 𝑘) − (3 𝑖 + 4 𝑗 + 1 𝑘) = 2 𝑖 − 3 𝑗 + 5 𝑘 ∴ 𝑟 = (3 𝑖 + 4 𝑗 + 1 𝑘) + 𝜆 (2 𝑖 − 3 𝑗 + 5 𝑘) Let the coordinates of the point where the line crosses the XY plane be (x, y, 0). So, 𝑟 = x 𝑖 + y 𝑗 + 0 𝑘 Since point crosses the plane, it will satisfy its equation Putting (2) in (1) x 𝑖 + y 𝑗 + 0 𝑘 = 3 𝑖 + 4 𝑗 + 1 𝑘 + 2𝜆 𝑖 − 3𝜆 𝑗 + 5𝜆 𝑘 = (3 + 2𝜆) 𝑖 + (4 − 3𝜆) 𝑗 + (1 + 5𝜆) 𝑘 Two vectors are equal if their corresponding components are equal So, Solving 0 = 1 + 5𝜆 ∴ 𝜆 = −𝟏𝟓 So, x = 3 + 𝜆 = 3 + 2 × −15 = 3 − 25 = 135 & y = 4 − 3𝜆 = 4 − 3 × −15 = 4 + 35 = 235 Therefore, the required coordinates are 𝟏𝟑𝟓, 𝟐𝟑𝟓,𝟎 Example 30 (Method 2) Find the coordinates of the point where the line through the points A (3, 4, 1) and B(5, 1, 6) crosses the XY-plane. The equation of a line passing through two points A( 𝑥1, 𝑦1, 𝑧1) and B( 𝑥2, 𝑦2, 𝑧2) is 𝒙 − 𝒙𝟏 𝒙𝟐 − 𝒙𝟏 = 𝒚 − 𝒚𝟏 𝒚𝟐 − 𝒚𝟏 = 𝒛 − 𝒛𝟏 𝒛𝟐 − 𝒛𝟏 Given the line passes through the points So, the equation of line is 𝑥 − 35 − 3 = 𝑦 − 41 − 4 = 𝑧 − 16 − 1 𝑥 − 32 = 𝑦 − 4−3 = 𝑧 − 15 = k So, Since, the line crosses the XY plane at (x, y, 0) z = 0 5k + 1 = 0 5k = −1 ∴ k = −𝟏𝟓 So, x = 2k + 3 = 2 × −15 + 3 = 3 − 25 = 135 y = −3 × −15 + 4 = 4 + 35 = 235 Therefore, the coordinates of the required point are 𝟏𝟑𝟓, 𝟐𝟑𝟓, 𝟎.

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Example 30 Important You are here

Chapter 11 Class 12 Three Dimensional Geometry

Serial order wise

About the Author

CA Maninder Singh

CA Maninder Singh is a Chartered Accountant for the past 8 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .